Question

Suppose 16.6 g of zinc acetate is dissolved in 350. mL of a 0.30 Maqueous solution of ammonium sulfate. Calculate the final molarity of zinc cation in the solution. You can assume the volume of the solution doesn't change when the zinc acetate is dissolved in it. Round your answer to 3 significant digits. Пм x 3 ?

Answer 1

2eacts uoiter Zincactate am as: Ammeni Suatees Zine Zine Sustute Acetate +2NHUCHCOO LAmmoiu Mass Number. Moles Molar Mass بمهممه numbern calculate we san ~\৪: zincactate Zn(chjcoo Zinc acetate As molar mass 183. 48 is 16. 6 133.48 lmol moles O.0905meles Melanit Numbos mles lute želution Velcine اسد moles 2elutiovn golute Lo No. meles CO 20M)x 350 ammeniu |00D

Becanse 1000 Thus 350 mU 1000 (o.20mal/L) (0.350 L) No. males mei kssdfats 0.1OSmales

Nou accond to tere balamced hemicas acetate acts sauatie mele zinc mele asittn anm We rase ammeniu O.1OS meles O1OS meRes react Jarith but thasra O.0905males Zinc acetare ase zinc acetate Tus zane acetate imiing actant Cee o.0aos meles acetare cansmed witer eactien ameniu s e would 2emain acetate tee selutien aele it cenveates to zinc 1 male News Zuson as tmele acetae Doonos mele ولنيط zine acetate O.0aO5 meles

Thus ourca حن )2)ح zins aes bolution the دعامطقيعنيع Zine Sulplnate dissowiates selution as Lmele Pxsبه moles of Z^ sou I mele then Zn+ O.0905 Znsom Praduca will males O.090s znz+ Thus ر طیر wممومی -O.O9Os catiens zine moles Meso Volume 350 ml lution- 350_L Nooo -O.350L. We caten zine as: De lanit moles zinc catien Velue okselution in l O 350 L o.258 melL O258M تحملم جنسم